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Math Help - Basic differential equations help

  1. #1
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    Basic differential equations help

    find the general solution for y''[x] = -y'[x]

    the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
    i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

    If anyone can help me, I would greatly appreciate it (ie worship them forever)

    Thanks

    ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...
    Last edited by lordpikachu; February 7th 2010 at 05:55 PM.
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  2. #2
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    Quote Originally Posted by lordpikachu View Post
    find the general solution for y''[x] = -y'[x]

    the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
    i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

    If anyone can help me, I would greatly appreciate it (ie worship them forever)

    Thanks

    ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...

    You are correct: observe that y''=-y\Longrightarrow \frac{y''}{y'}=-1\Longrightarrow \ln y'=-x+C\Longrightarrow y'=Ke^{-x}\Longrightarrow y=Ke^{-x} , integrating twice.

    Tonio
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  3. #3
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    Quote Originally Posted by lordpikachu View Post
    find the general solution for y''[x] = -y'[x]

    the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
    i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?

    If anyone can help me, I would greatly appreciate it (ie worship them forever)

    Thanks

    ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...
    set up the associated characteristic polynomial and solve for the roots. those roots are what appear in the exponents of the exponentials above.
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  4. #4
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    Quote Originally Posted by tonio View Post
    You are correct: observe that y''=-y\Longrightarrow \frac{y''}{y'}=-1\Longrightarrow \ln y'=-x+C\Longrightarrow y'=Ke^{-x}\Longrightarrow y=Ke^{-x} , integrating twice.

    Tonio
    This works. I get  y=c1 + c2e^{-x}
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  5. #5
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    OHHH i get it

    Oh wow thanks!! I LOVE YOU GUYS. // thanks a million times.
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  6. #6
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    Quote Originally Posted by lordpikachu View Post
    find the general solution for y''[x] = -y'[x]

    the answer is y=c1e^(tx1) + c2e^(tx2) where c1, c2, x1, x2 are separate variables/constants...
    i don't understand..i keep getting that y=ce^-x but that comes nowhere close to what the real answer is?
    The general solution is, as you have already been told, y= c_1+ c_2e^{-t} (it is your choice to use t or x as the independent variable since it is not given in the equation). My point is that this is exactly of the form c_1e^{tx_1}+ c_2e^{tx_2} with x_1= 0 and x_2= -1.

    If anyone can help me, I would greatly appreciate it (ie worship them forever)
    We will forward you an address to send your tithes to. You will have to purchase your own white robes.

    Thanks

    ps. should this go in the diffeq forum? i wasn't sure because this is basic diffeq you learn in calc...
    Follow Math Help Forum on Facebook and Google+

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